No,
we’re not yet finished with these exotic mathematical denizens! We want to give
yet more examples, and especially showing different ways to obtain the Laurent
equivalent of different functions.
Then:
(2u) 4 / 4! + …….
= e 2 / (z
- 1)3 + 2 e 2 / (z
- 1)2 + 2 e 2 / (z
– 1) +
4 e 2 / 3 + 2 e 2 / 3 (z – 1) + …….
4 e 2 / 3 + 2 e 2 / 3 (z – 1) + …….
Ex. (2): f(z) = z – sin z /
z 3
sin(z)
= z - z3 / 3! + z5
/ 5! …..
- (-1) n- 1 z 2n
-1/ (2n -1)! +
Then:
z – sin z / z 3 = 1/
z 3 [z -
(z - z3 / 3! + z5
/ 5! …..
) ]
= 1/ z 3 [z3 / 3! - z5 / 5! + z 7
/ 7! -
……….
= 1/3! - z 2 / 5! + z 4 / 7! - z 6 / 9! + ……..
Ex. (3)
Using
the sine series we can write:
sin (4z) / z 4 = 1/ z 4 [4z - 4z3 / 3! + 4 z5
/ 5! +
……..]
= 4 / z 3 - 43 / 3! z + 45 z /
5! +
…..
= 4 z -3 - 43
z -1
/ 3! + 45 z /
5! +
…..
Which
on inspection is found to have the form:
å¥ n = 0 (-1)
n 4 2n+
1 z 2n -3 / (2n + 1)!
Check term number 3 (n = 2):
4 2n+ 1 = 4 2(2)+ 1 =
45
z 2n -3 = z 2(2) - 3 = z 1= z
(2n
+ 1)! = (2(2) + 1)!
= 5!
Then: we get 45 z / 5!
A
removable singularity occurs at z = 0 so we expect the region of convergence to
be:
÷ z ÷ >
0
Problems for Math Mavens:
Let
f(z) = exp (-1/z 2) / z5
a)
Show the Laurent series can be written:
å¥ n = 0 (-1)
n / n ! z 2n
+ 5
b)Specify where the singularity would occur and the precise region of convergence
Challenge
Problem:
For über Math Mavens only –
Let
f(z) = 7 z 2 + 9 z - 18 /
z 3 - 9z
Find
Laurent series for the convergence regions:
a)
0 < ÷ z ÷ <
3 and
b) ÷ z ÷ >
3
(Hint:
Approach the function f(z) by resort to partial fraction decomposition)
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